Calculate the wavenumber for the longest wavelength transition in balmer series of atomic hydrogen.

The correct option is C $1.52\times {10}^4{\text{cm}}^{-1}$
For longest wavelength transition in balmer series,
$n_1=2$
$n_2=3$
According to Rydberg equation
${\text{Wavenumber = R}}_h\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$=109677\times (\frac{1}{2^2}-\frac{1}{3^2})$
$=1.52\times {10}^4{\text{cm}}^{-1}$